A generalization on weak contractions in partially ordered b-metric spaces and its application to quadratic integral equations

Abstract

We introduce the notion of almost generalized $(\psi ,\phi ,L)$-contractive mappings, and establish the coincidence and common fixed point results for this class of mappings in partially ordered complete b-metric spaces. Our results extend and improve several known results from the context of ordered metric spaces to the setting of ordered b-metric spaces. As an application, we prove the existence of a unique solution to a class of nonlinear quadratic integral equations.

1 Introduction

Fixed points theorems in partially ordered metric spaces were firstly obtained in 2004 by Ran and Reurings [1], and then by Nieto and Rodríguez-López [2]. In this direction several authors obtained further results under weak contractive conditions (see, e.g., [3–8]). Berinde initiated in [9] the concept of almost contractions and obtained several interesting fixed point theorems. This has been a subject of intense study since then; see, e.g., [10–20]. Some authors used related notions as ‘condition (B)’ (Babu et al. [21]) and ‘almost generalized contractive condition’ for two maps (Ćirić et al. [22]), and for four maps (Aghajani et al. [23]). See also a note by Pacurar [15]. On the other hand, the concept of b-metric space was introduced by Czerwik in [24]. After that, several interesting results of the existence of fixed point for single-valued and multivalued operators in b-metric spaces have been obtained (see [25–40]). Pacurar [41] proved some results on sequences of almost contractions and fixed points in b-metric spaces. Recently, Hussain and Shah [42] obtained results on KKM mappings in cone b-metric spaces. Using the concepts of partially ordered metric spaces, almost generalized contractive condition, and b-metric spaces, we define a new concept of almost generalized $(\psi ,\phi ,L)$-contractive condition. In this paper, some coincidence and common fixed point theorems for mappings satisfying almost generalized $(\psi ,\phi ,L)$-contractive condition in the setup of partially ordered complete b-metric spaces are proved. Consistent with [43] and [[40], p.264], the following definitions and results will be needed in the sequel.

Definition 1.1 [43]

Let X be a (nonempty) set and $s\ge 1$ be a given real number. A function $d:X\times X\to {\mathbb{R}}^{+}$ is said to be a b-metric space iff for all $x,y,z\in X$, the following conditions are satisfied:

• (i) $d(x,y)=0$ iff $x=y$,

• (ii) $d(x,y)=d(y,x)$,

• (iii) $d(x,y)\le s[d(x,z)+d(z,y)]$.

The pair $(X,d)$ is called a b-metric space with the parameter s.

It should be noted that the class of b-metric spaces is effectively larger than that of metric spaces, since a b-metric is a metric, when $s=1$.

The following example shows that in general a b-metric does not necessarily need to be a metric (see, also, [40]).

Example 1.1 [44]

Let $(X,d)$ be a metric space and $\rho (x,y)={(d(x,y))}^p$, where $p>1$ is a real number. Then ρ is a b-metric with $s=2^{p-1}$. However, if $(X,d)$ is a metric space, then $(X,\rho )$ is not necessarily a metric space. For example, if $X=\mathbb{R}$ is the set of real numbers and $d(x,y)=|x-y|$ is the usual Euclidean metric, then $\rho (x,y)={(x-y)}^s$ is a b-metric on ℝ with $s=2$, but it is not a metric on ℝ.

Also, the following example of a b-metric space is given in [45].

Example 1.2 [45]

Let X be the set of Lebesgue measurable functions on $[0,1]$ such that ${\int }_0^1{|f(x)|}^{2}dx<\mathrm{\infty }$. Define $D:X\times X\to [0,\mathrm{\infty })$ by $D(f,g)={\int }_0^1{|f(x)-g(x)|}^{2}dx$. As ${({\int }_0^1{|f(x)-g(x)|}^{2}dx)}^{\frac{1}{2}}$ is a metric on X, then, from the previous example, D is a b-metric on X, with $s=2$, where the b-metric D is defined with $D(x,y)=\parallel d(x,y)\parallel$, d is a cone metric (also see [46–49]).

Khamsi [50] also showed that each cone metric space over a normal cone has a b-metric structure.

Definition 1.2 [6]

We shall say that the mapping T is g-nondecreasing if

$$gx\le gy⟹Tx\le Ty.$$

2 Main results

Throughout the paper, let Ψ be the family of all functions $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ satisfying the following conditions:

• (a) ψ is continuous,

• (b) ψ is nondecreasing,

• (c) $\psi (0)=0<\psi (t)$ for every $t>0$.

We denote by Φ the set of all functions $\phi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ satisfying the following conditions:

• (i) φ is right continuous,

• (ii) φ is nondecreasing,

• (iii) $\phi (t)<t$ for every $t>0$.

Let $(X,d,\le )$ be a partially ordered b-metric space and $T:X\to X$ and $g:X\to X$ be two mappings. Set

$$M(x,y)=max\{d(gx,gy),d(gx,Tx),d(gy,Ty),\frac{d(gx,Ty)+d(gy,Tx)}{2s}\}$$

and

$$N(x,y)=min\{d(gx,Tx),d(gy,Ty),d(gx,Ty),d(gy,Tx)\}.$$

Now, we introduce the following definition.

Definition 2.1 Let $(X,d,\le )$ be a partially ordered b-metric space. We say that $T:X\to X$ is an almost generalized $(\psi ,\phi ,L)$-contractive mapping with respect to $g:X\to X$ for some $\psi \in \mathrm{\Psi }$, $\phi \in \mathrm{\Phi }$, and $L\ge 0$ if

$$\psi (s^{3}d(Tx,Ty))\le \phi \left(\psi (M(x,y))\right)+L\psi (N(x,y))$$

(2.1)

for all $x,y\in X$ with $gx\le gy$.

Now, we establish some results for the existence of coincidence point and common fixed point of mappings satisfying almost generalized $(\psi ,\phi ,L)$-contractive condition in the setup of partially ordered b-metric spaces. The first result in this paper is the following coincidence point theorem.

Theorem 2.1 Suppose that $(X,d,\le )$ is a partially ordered complete b-metric space. Let $T:X\to X$ be an almost generalized $(\psi ,\phi ,L)$-contractive mapping with respect to $g:X\to X$, and T and g are continuous such that T is a monotone g-nondecreasing mapping, commutative with g and $T(X)\subseteq g(X)$. If there exists $x_0\in X$ such that $g{x}_0\le T{x}_0$, then T and g have a coincidence point in X.

Proof By the given assumptions, there exists $x_0\in X$ such that $g{x}_0\le T{x}_0$. Since $T(X)\subseteq g(X)$, we can define $x_1\in X$ such that $g{x}_1=T{x}_0$, then $g{x}_0\le T{x}_0=g{x}_1$. Also there exists $x_2\in X$ such that $g{x}_2=T{x}_1$. Since T is a monotone g-nondecreasing mapping, we have

$$g{x}_1=T{x}_0\le T{x}_1=g{x}_2.$$

Continuing in this way, we construct a sequence $\{x_n\}$ in X such that for all $n=0,1,2,\dots$ ,

$$g{x}_{n+1}=T{x}_n$$

(2.2)

for which

$$g{x}_0\le g{x}_1\le g{x}_2\le \cdots \le g{x}_n\le g{x}_{n+1}\le \cdots .$$

(2.3)

If there exists $k_0\in \mathbb{N}$ such that $g{x}_{k_0+1}=g{x}_{k_0}$, then $g{x}_{k_0}=T{x}_{k_0}$. This means that $x_{k_0}$ is a coincidence point of T, g, and the proof is finished. Thus, $g{x}_{n+1}\ne g{x}_n$ for all $n\in \mathbb{N}$. From (2.2) and (2.3) and the inequality (2.1) with $(x,y)=(x_n,x_{n+1})$, we have

$$\begin{array}{rl}\psi (d(g{x}_{n+1},g{x}_{n+2}))& \le \psi (s^{3}d(g{x}_{n+1},g{x}_{n+2}))=\psi (s^{3}d(T{x}_n,T{x}_{n+1}))\\ \le \phi \left(\psi (M(x_n,x_{n+1}))\right)+L\psi (N(x_n,x_{n+1})),\end{array}$$

(2.4)

where

$$\begin{array}{rcl}M(x_n,x_{n+1})& =& max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_n,T{x}_n),d(g{x}_{n+1},T{x}_{n+1}),\\ \frac{d(g{x}_n,T{x}_{n+1})+d(g{x}_{n+1},T{x}_n)}{2s}\}\\ =& max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2}),\frac{d(g{x}_n,g{x}_{n+2})}{2s}\}\end{array}$$

and

$$N(x_n,x_{n+1})=min\{d(g{x}_n,T{x}_n),d(g{x}_{n+1},T{x}_{n+1}),d(g{x}_n,T{x}_{n+1}),d(g{x}_{n+1},T{x}_n)\}=0.$$

Since

$$\frac{d(g{x}_n,g{x}_{n+2})}{2s}\le \frac{d(g{x}_n,g{x}_{n+1})+d(g{x}_{n+1},g{x}_{n+2})}{2}\le max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2})\},$$

then we get

$$\begin{array}{r}M(x_n,x_{n+1})=max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2})\},\\ N(x_n,x_{n+1})=0.\end{array}$$

(2.5)

By (2.4) and (2.5), we have

$$\psi (d(g{x}_{n+1},g{x}_{n+2}))\le \phi \left(\psi (max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2})\})\right).$$

(2.6)

Suppose that $max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2})\}=d(g{x}_{n+1},g{x}_{n+2})>0$ for some $n\in \mathbb{N}$, then by (2.6)

$$\psi (d(g{x}_{n+1},g{x}_{n+2}))\le \phi \left(\psi (d(g{x}_{n+1},g{x}_{n+2}))\right)<\psi (d(g{x}_{n+1},g{x}_{n+2}));$$

a contradiction. Hence,

$$max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2})\}=d(g{x}_n,g{x}_{n+1})$$

and thus

$$\psi (d(g{x}_{n+1},g{x}_{n+2}))\le \phi \left(\psi (d(g{x}_n,g{x}_{n+1}))\right)<\psi (d(g{x}_n,g{x}_{n+1})).$$

Thus, we get

$$\psi (d(g{x}_{n+1},g{x}_{n+2}))<\psi (d(g{x}_n,g{x}_{n+1}))$$

for all $n\in \mathbb{N}$. Now, from

$$\begin{array}{rl}\psi (d(g{x}_n,g{x}_{n+1}))& \le \phi \left(\psi (d(g{x}_{n-1},g{x}_n))\right)\le {\phi }^2\left(\psi (d(g{x}_{n-2},g{x}_{n-1}))\right)\\ \le \cdots \le {\phi }^n\left(\psi (d(g{x}_0,g{x}_1))\right)\end{array}$$

and the property of φ, we obtain ${lim}_{n\to \mathrm{\infty }}\psi (d(g{x}_n,g{x}_{n+1}))=0$, and consequently

$$\underset{n\to \mathrm{\infty }}{lim}d(g{x}_n,g{x}_{n+1})=0.$$

(2.7)

Now, we shall prove that $\{g{x}_n\}$ is a Cauchy sequence in $(X,d)$. Suppose, on the contrary, that $\{g{x}_n\}$ is not a Cauchy sequence. Then there exist $ϵ>0$ and subsequences $\{g{x}_{m(k)}\}$, $\{g{x}_{n(k)}\}$ of $\{g{x}_n\}$ with $m(k)>n(k)\ge k$ such that

$$d(g{x}_{n(k)},g{x}_{m(k)})\ge ϵ.$$

(2.8)

Additionally, corresponding to $n(k)$, we may choose $m(k)$ such that it is the smallest integer satisfying (2.8) and $m(k)>n(k)\ge k$. Thus,

$$d(g{x}_{n(k)},g{x}_{m(k)-1})<ϵ.$$

(2.9)

Using the triangle inequality in b-metric space and (2.8) and (2.9) we obtain

$$\begin{array}{rl}ϵ& \le d(g{x}_{m(k)},g{x}_{n(k)})\le sd(g{x}_{m(k)},g{x}_{m(k)-1})+sd(g{x}_{m(k)-1},g{x}_{n(k)})\\ <sd(g{x}_{m(k)},g{x}_{m(k)-1})+sϵ.\end{array}$$

Taking the upper limit as $k\to \mathrm{\infty }$ and using (2.7) we obtain

$$ϵ\le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{n(k)},g{x}_{m(k)})\le sϵ.$$

(2.10)

Also

$$\begin{array}{rl}ϵ& \le d(g{x}_{n(k)},g{x}_{m(k)})\le sd(g{x}_{n(k)},g{x}_{m(k)+1})+sd(g{x}_{m(k)+1},g{x}_{m(k)})\\ \le s^{2}d(g{x}_{n(k)},g{x}_{m(k)})+s^{2}d(g{x}_{m(k)},g{x}_{m(k)+1})+sd(g{x}_{m(k)+1},g{x}_{m(k)})\\ \le s^{2}d(g{x}_{n(k)},g{x}_{m(k)})+(s^2+s)d(g{x}_{m(k)},g{x}_{m(k)+1}).\end{array}$$

So from (2.7) and (2.10), we have

$$\frac{ϵ}{s}\le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{n(k)},g{x}_{m(k)+1})\le s^2ϵ.$$

(2.11)

Also

$$\begin{array}{rl}ϵ& \le d(g{x}_{m(k)},g{x}_{n(k)})\le sd(g{x}_{m(k)},g{x}_{n(k)+1})+sd(g{x}_{n(k)+1},g{x}_{n(k)})\\ \le s^{2}d(g{x}_{m(k)},g{x}_{n(k)})+s^{2}d(g{x}_{n(k)},g{x}_{n(k)+1})+sd(g{x}_{n(k)+1},g{x}_{n(k)})\\ \le s^{2}d(g{x}_{m(k)},g{x}_{n(k)})+(s^2+s)d(g{x}_{n(k)},g{x}_{n(k)+1}).\end{array}$$

So from (2.7) and (2.10), we have

$$\frac{ϵ}{s}\le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{m(k)},g{x}_{n(k)+1})\le s^2ϵ.$$

(2.12)

Also

$$d(g{x}_{n(k)+1},g{x}_{m(k)})\le sd(g{x}_{n(k)+1},g{x}_{m(k)+1})+sd(g{x}_{m(k)+1},g{x}_{m(k)}),$$

so from (2.7) and (2.12), we have

$$\frac{ϵ}{s^2}\le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{n(k)+1},g{x}_{m(k)+1}).$$

(2.13)

Linking (2.7), (2.10), (2.11) together with (2.12) we get

$$\begin{array}{c}\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_{n(k)},x_{m(k)})\hfill \\ =max\{\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{n(k)},g{x}_{m(k)}),\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{n(k)},g{x}_{n(k)+1}),\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{m(k)},g{x}_{m(k)+1}),\hfill \\ \frac{{lim\hspace{0.17em}sup}_{k\to \mathrm{\infty }}d(g{x}_{n(k)},g{x}_{m(k)+1})+{lim\hspace{0.17em}sup}_{k\to \mathrm{\infty }}d(g{x}_{m(k)},g{x}_{n(k)+1})}{2s}\}\hfill \\ \le max\{sϵ,0,0,\frac{s^2ϵ+s^2ϵ}{2s}\}=sϵ.\hfill \end{array}$$

So,

$$\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_{n(k)},x_{m(k)})\le ϵs.$$

(2.14)

Similarly, we have

$$\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}N(x_{n(k)},x_{m(k)})=0.$$

(2.15)

Since $m(k)>n(k)$ from (2.2), we have

$$g{x}_{n(k)}\le g{x}_{m(k)}.$$

Thus,

$$\begin{array}{rl}\psi (s^{3}d(g{x}_{n(k)+1},g{x}_{m(k)+1}))& =\psi (s^{3}d(T{x}_{n(k)},T{x}_{m(k)}))\\ \le \phi \left(\psi (M(x_{n(k)},x_{m(k)}))\right)+L\psi (N(x_{n(k)},x_{m(k)})).\end{array}$$

Passing to the upper limit as $k\to \mathrm{\infty }$, and using (2.13), (2.14), and (2.15), we get

$$\begin{array}{rl}\psi (sϵ)& \le \psi (s^3\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{n(k)+1},g{x}_{m(k)+1}))=\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi (s^{3}d(g{x}_{n(k)+1},g{x}_{m(k)+1}))\\ =\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi (s^{3}d(T{x}_{n(k)},T{x}_{m(k)}))\\ \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\phi \left(\psi (M(x_{n(k)},x_{m(k)}))\right)+\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}L\psi (N(x_{n(k)},x_{m(k)}))\\ =\phi \left(\psi (\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_{n(k)},x_{m(k)}))\right)+L\psi (\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}N(x_{n(k)},x_{m(k)}))\\ \le \phi (\psi (ϵs))<\psi (sϵ),\end{array}$$

which is a contradiction. Thus, we proved that $\{g{x}_n\}$ is a Cauchy sequence in $(X,d)$. Since X is a complete b-metric space, there exists $x\in X$ such that

$$\underset{n\to \mathrm{\infty }}{lim}g{x}_{n+1}=x.$$

(2.16)

From the commutativity of T and g, we have

$$g(g{x}_{n+1})=g(T(x_n))=T(g{x}_n).$$

(2.17)

Letting $n\to \mathrm{\infty }$ in (2.17) and from the continuity of T and g, we get

$$gx=\underset{n\to \mathrm{\infty }}{lim}g(g{x}_{n+1})=\underset{n\to \mathrm{\infty }}{lim}T(g{x}_n)=T\left(\underset{n\to \mathrm{\infty }}{lim}g{x}_n\right)=T(x).$$

This implies that x is a coincidence point of T and g. This completes the proof. □

Now, we will prove the following result.

Theorem 2.2 Suppose that $(X,d,\le )$ is a partially ordered complete b-metric space. Let $T:X\to X$ be an almost generalized $(\psi ,\phi ,L)$-contractive mapping with respect to $g:X\to X$, T is a g-nondecreasing mapping and $T(X)\subseteq g(X)$. Also suppose

$$\begin{array}{r}\mathit{\text{if }}\{g{x}_n\}\subset X\mathit{\text{ is a nondecreasing sequence with }}g{x}_n\to gz\mathit{\text{ in }}gX,\\ \mathit{\text{then }}g{x}_n\le gz,gz\le g(gz)\mathrm{\forall }n\mathit{\text{ hold}}.\end{array}$$

(2.18)

Also suppose gX is closed. If there exists $x_0\in X$ such that $g{x}_0\le T{x}_0$, then T and g have a coincidence. Further, if T and g commute at their coincidence points, then T and g have a common fixed point.

Proof As in the proof of Theorem 2.1, we can show that $\{g{x}_n\}$ is a Cauchy sequence. Since gX is a closed, there exists $x\in X$ such that

$$\underset{n\to \mathrm{\infty }}{lim}g{x}_{n+1}=gx.$$

(2.19)

Now we show that x is a coincidence point of T and g. Since from (2.18) and (2.19) we have $g{x}_n\le gx$ for all n, then by the triangle inequality in a b-metric space and (2.1), we get

$$\begin{array}{c}d(gx,Tx)\le sd(gx,g{x}_{n+1})+sd(g{x}_{n+1},Tx)=sd(gx,g{x}_{n+1})+sd(T{x}_n,Tx),\hfill \\ \psi (d(gx,Tx))\le \underset{n\to \mathrm{\infty }}{lim}\psi (sd(T{x}_n,Tx))\le \underset{n\to \mathrm{\infty }}{lim}\psi (s^{3}d(T{x}_n,Tx))\hfill \\ \phantom{\psi (d(gx,Tx))}\le \underset{n\to \mathrm{\infty }}{lim}[\phi \left(\psi (M(x_n,x))\right)+L\psi (N(x_n,x))]\hfill \\ \phantom{\psi (d(gx,Tx))}\le \phi \left(\psi (d(gx,Tx))\right)<\psi (d(gx,Tx)).\hfill \end{array}$$

Indeed,

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}M(x_n,x)& =\underset{n\to \mathrm{\infty }}{lim}max\{d(g{x}_n,gx),d(g{x}_n,T{x}_n),d(gx,Tx),\frac{d(g{x}_n,Tx)+d(gx,T{x}_n)}{2s}\}\\ =d(gx,Tx)\end{array}$$

and

$$\underset{n\to \mathrm{\infty }}{lim}N(x_n,x)=\underset{n\to \mathrm{\infty }}{lim}min\{d(g{x}_n,T{x}_n),d(gx,Tx),d(g{x}_n,Tx),d(gx,T{x}_n)\}=0.$$

Hence $d(gx,Tx)=0$, that is, $Tx=gx$. Thus we proved that T and g have a coincidence. Suppose now that T and g commute at x. Set $y=Tx=gx$. Then

$$Ty=T(gx)=g(Tx)=gy.$$

Since from (2.18) we have $gx\le g(gx)=gy$ and as $gx=Tx$ and $gy=Ty$, from (2.1) we obtain

$$\begin{array}{rcl}\psi (d(Tx,Ty))& \le & \psi (s^{3}d(Tx,Ty))\le \phi \left(\psi (M(x,y))\right)+L\psi (N(x,y))\\ =& \phi \left(\psi (max\{d(gx,gy),d(gx,Tx),d(gy,Ty),\frac{d(gx,Ty)+d(gy,Tx)}{2s}\})\right)\\ +L\psi (min\{d(gx,Tx),d(gy,Ty),d(gx,Ty),d(gy,Tx)\})\\ =& \phi \left(\psi (d(Tx,Ty))\right)<\psi (d(Tx,Ty)).\end{array}$$

Hence $d(Tx,Ty)=0$, that is, $y=Tx=Ty$. Therefore, $Ty=gy=y$. Thus we proved that T and g have a common fixed point. □

In the following, we deduce some fixed point theorems from our main results given by Theorems 2.1 and 2.2.

Corollary 2.3 Let $(X,d,\le )$ be a partially ordered complete b-metric space and $T:X\to X$ is a nondecreasing mapping. Suppose there exist $\psi \in \mathrm{\Psi }$, $\phi \in \mathrm{\Phi }$, and $L\ge 0$ such that

$$\psi (s^{3}d(Tx,Ty))\le \phi \left(\psi (M(x,y))\right)+L\psi (N(x,y)),$$

where

$$M(x,y)=max\{d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2s}\}$$

and

$$N(x,y)=min\{d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}$$

for all $x,y\in X$ with $x\le y$. Also suppose either

• (a) if $\{x_n\}\subset X$ is a nondecreasing sequence with $x_n\to z$ in X, then $x_n\le z$, for all n, holds, or

• (b) T is continuous.

If there exists $x_0\in X$ such that $x_0\le T{x}_0$, then T has a fixed point in X.

Example 2.1 Let X be the set of Lebesgue measurable functions on $[0,1]$ such that ${\int }_0^1|x(t)|dt<\mathrm{\infty }$. Define $D:X\times X\to [0,\mathrm{\infty })$ by

$$D(x,y)={\left({\int }_0^1\right|x(t)-y(t)\left|dt\right)}^2.$$

Then D is a b-metric on X, with $s=2$. Also, this space can also be equipped with a partial order given by

$$x,y\in X,x⪯y⟺x(t)\le y(t)\text{for any }t\in [a,b].$$

The operator $T:X\to X$ defined by

$$Tx(t)=t^n+e^t+\frac{\sqrt{2}}{4}ln\left(\right|x(t)|+1).$$

(2.20)

Now, we prove that T has a fixed point. For all $x,y\in X$ with $x\le y$, we have

$$\begin{array}{rcl}\sqrt{2^{3}D(Tx,Ty)}& =& \sqrt{2^3{\left({\int }_0^1\right|Tx(t)-Ty(t)\left|dt\right)}^2}\\ \le & 2\sqrt{2}{\int }_0^1|\frac{\sqrt{2}}{4}ln\left(\right|x(t)|+1)-\frac{\sqrt{2}}{4}ln\left(\right|y(t)|+1)|dt\\ \le & {\int }_0^1\left|(ln\left(\right|x(t)|+1)-ln\left(\right|y(t)|+1))\right|dt\\ \le & {\int }_0^{1}ln\left(\frac{|x(t)|+1}{|y(t)|+1}\right)dt\\ \le & {\int }_0^{1}ln(1+\frac{|x(t)-y(t)|}{|y(t)|+1})dt\\ \le & ln(1+{\int }_0^1|x(t)-y(t)\left|dt\right)\\ \le & ln(1+\sqrt{{\left({\int }_0^1\right|x(t)-y(t)\left|dt\right)}^2})\\ \le & ln(1+\sqrt{D(x,y)}).\end{array}$$

Now, if we define $\phi (t)=ln(1+t)$, $\psi (t)=\sqrt{t}$, and $x_0=0$. Thus, by Corollary 2.3 we see that T has a fixed point.

Remark 2.1 Corollary 2.3 extends and generalizes many existing fixed point theorems in the literature [2, 3, 51, 52].

The following result is the immediate consequence of Corollary 2.3.

Corollary 2.4 Let $(X,d,\le )$ be a partially ordered complete b-metric space and $T:X\to X$ is a nondecreasing mapping. Suppose there exists $\phi \in \mathrm{\Phi }$ such that

$$s^{3}d(Tx,Ty)\le \phi (max\{d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2s}\})$$

(2.21)

for all $x,y\in X$ with $x\le y$. Also suppose either

• (a) if $\{x_n\}\subset X$ is a nondecreasing sequence with $x_n\to z$ in X, then $x_n\le z$, for all n, holds, or

• (b) T is continuous.

If there exists $x_0\in X$ such that $x_0\le T{x}_0$, then T has a fixed point in X.

Remark 2.2 Corollary 2.4 is a generalization to [[3], Theorem 1.3].

Taking $\phi (t)=\lambda t$, $0<\lambda <1$, in Corollary 2.4 we obtain the following generalization of the results in [1, 53].

Corollary 2.5 Let $(X,d,\le )$ be a partially ordered complete b-metric space and $T:X\to X$ is a nondecreasing mapping. Suppose there exists $\phi \in \mathrm{\Phi }$ such that

$$s^{3}d(Tx,Ty)\le \lambda max\{d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2s}\}$$

for all $x,y\in X$ with $x\le y$. Also suppose either

• (a) if $\{x_n\}\subset X$ is a nondecreasing sequence with $x_n\to z$ in X, then $x_n\le z$, for all n, holds, or

• (b) T is continuous.

If there exists $x_0\in X$ such that $x_0\le T{x}_0$, then T has a fixed point in X.

Corollary 2.6 Let $(X,d,\le )$ be a partially ordered complete b-metric space and $T:X\to X$ is a nondecreasing mapping. Suppose there exist $\psi \in \mathrm{\Psi }$ and $0\le \lambda <1$ such that

$$\psi (s^{3}d(Tx,Ty))\le \lambda \psi (d(x,y))$$

for all $x,y\in X$ with $x\le y$. Also suppose either

• (a) if $\{x_n\}\subset X$ is a nondecreasing sequence with $x_n\to z$ in X, then $x_n\le z$, for all n, holds, or

• (b) T is continuous.

If there exists $x_0\in X$ such that $x_0\le T{x}_0$, then T has a fixed point in X.

3 Application to integral equations

Here, in this section, we wish to study the existence of a unique solution to a nonlinear quadratic integral equation, as an application to the our fixed point theorem. Consider the integral equation

$$x(t)=h(t)+\lambda {\int }_0^{1}k(t,s)f(s,x(s))ds,t\in I=[0,1],\lambda \ge 0.$$

(3.1)

Let Γ denote the class of those functions $\gamma :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ for which $\gamma \in \mathrm{\Phi }$ and ${(\gamma (t))}^p\le \gamma (t^p)$, for all $p\ge 1$.

For example, ${\gamma }_1(t)=kt$, where $0\le k<1$ and ${\gamma }_2(t)=\frac{t}{t+1}$ are in Γ.

We will analyze (3.1) under the following assumptions:

(a₁) $f:I\times \mathbb{R}\to \mathbb{R}$ is continuous monotone nondecreasing in x, $f(t,x)\ge 0$ and there exist constant $0\le L<1$ and $\gamma \in \mathrm{\Gamma }$ such that for all $x,y\in \mathbb{R}$ and $x\ge y$

$$|f(t,x)-f(t,y)|\le L\gamma (x-y).$$

(a₂) $h:I\to \mathbb{R}$ is a continuous function.

(a₃) $k:I\times I\to \mathbb{R}$ is continuous in $t\in I$ for every $s\in I$ and measurable in $s\in I$ for all $t\in I$ such that

$${\int }_0^{1}k(t,s)ds\le K$$

and $k(t,s)\ge 0$.

(a₄) There exists $\alpha \in C(I)$ such that

$$\alpha (t)\le h(t)+\lambda {\int }_0^{1}k(t,s)f(s,\alpha (s))ds.$$

(a₅) $L^p{\lambda }^p{K}^p\le \frac{1}{2^{3p-3}}$.

We consider the space $X=C(I)$ of continuous functions defined on $I=[0,1]$ with the standard metric given by

$$\rho (x,y)=\underset{t\in I}{sup}|x(t)-y(t)|\text{for }x,y\in C(I).$$

This space can also be equipped with a partial order given by

$$x,y\in C(I),x\le y⟺x(t)\le y(t)\text{for any }t\in I.$$

Now for $p\ge 1$, we define

$$d(x,y)={(\rho (x,y))}^p={\left(\underset{t\in I}{sup}\right|x(t)-y(t)\left|\right)}^p=\underset{t\in I}{sup}|x(t)-y(t){|}^p\text{for }x,y\in C(I).$$

It is easy to see that $(X,d)$ is a complete b-metric space with $s=2^{p-1}$ [44].

For any $x,y\in X$ and each $t\in I$, $max\{x(t),y(t)\}$ and $min\{x(t),y(t)\}$ belong to X and are upper and lower bounds of x, y, respectively. Therefore, for every $x,y\in X$, one can take $max\{x,y\},min\{x,y\}\in X$ which are comparable to x, y. Now, we formulate the main result of this section.

Theorem 3.1 Under assumptions (a₁)-(a₅), (3.1) has a unique solution in $C(I)$.

Proof We consider the operator $T:X\to X$ defined by

$$T(x)(t)=h(t)+\lambda {\int }_0^{1}k(t,s)f(s,x(s))ds\text{for }t\in I.$$

By virtue of our assumptions, T is well defined (this means that if $x\in X$ then $T(x)\in X$). For $x\le y$, and $t\in I$ we have

$$\begin{array}{rl}T(x)(t)-T(y)(t)& =h(t)+\lambda {\int }_0^{1}k(t,s)f(s,x(s))ds-h(t)-\lambda {\int }_0^{1}k(t,s)f(s,y(s))ds\\ =\lambda {\int }_0^{1}k(t,s)[f(s,x(s))-f(s,y(s))]ds\le 0.\end{array}$$

Therefore, T has the monotone nondecreasing property. Also, for $x\le y$, we have

$$\begin{array}{rl}|T(x)(t)-T(y)(t)|& =|h(t)+\lambda {\int }_0^{1}k(t,s)f(s,x(s))ds-h(t)-\lambda {\int }_0^{1}k(t,s)f(s,y(s))ds|\\ \le \lambda {\int }_0^{1}k(t,s)|f(s,x(s))-f(s,y(s))|ds\\ \le \lambda {\int }_0^{1}k(t,s)L\gamma (y(s)-x(s))ds.\end{array}$$

Since the function γ is nondecreasing and $x\le y$, we have

$$\gamma (y(s)-x(s))\le \gamma \left(\underset{t\in I}{sup}\right|x(s)-y(s)\left|\right)=\gamma (\rho (x,y)),$$

hence

$$|T(x)(t)-T(y)(t)|\le \lambda {\int }_0^{1}k(t,s)L\gamma (\rho (x,y))ds\le \lambda KL\gamma (\rho (x,y)).$$

Then we obtain

$$\begin{array}{rl}d(T(x),T(y))& =\underset{t\in I}{sup}|T(x)(t)-T(y)(t){|}^p\\ \le {\left\{\lambda KL\gamma (\rho (x,y))\right\}}^p={\lambda }^p{K}^p{L}^p\gamma {(\rho (x,y))}^p\\ \le {\lambda }^p{K}^p{L}^p\gamma \left(\rho {(x,y)}^p\right)={\lambda }^p{K}^p{L}^p\gamma (d(x,y))\\ \le {\lambda }^p{K}^p{L}^p\phi (max\{d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2s}\})\\ \le \frac{1}{2^{3p-3}}\phi (max\{d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2s}\}).\end{array}$$

This proves that the operator T satisfies the contractive condition (2.21) appearing in Corollary 2.4. Also, let α, β be the functions appearing in assumption (a₄); then, by (a₄), we get $\alpha \le T(\alpha )$. So, (3.1) has a solution and the proof is complete. □

Example 3.1 Consider the following functional integral equation:

$$x(t)=\frac{t^2}{1+t^4}+\frac{1}{27}{\int }_0^1\frac{e^{-s}sint}{2(1+t)}\frac{|x(s)|}{1+|x(s)|}ds$$

(3.2)

for $t\in [0,1]$. Observe that this equation is a special case of (3.1) with

$$\begin{array}{c}h(t)=\frac{t^2}{1+t^4},\hfill \\ k(t,s)=\frac{e^{-s}}{1+t},\hfill \\ f(t,x)=\frac{sint}{2}\frac{|x|}{1+|x|}.\hfill \end{array}$$

Indeed, by using $\gamma (t)=\frac{1}{3}t$ we see that $\gamma \in \mathrm{\Phi }$ and ${(\gamma (t))}^p={(\frac{1}{3}t)}^p=\frac{1}{3^p}{t}^p\le \frac{1}{3}{t}^p=\gamma (t^p)$, for all $p\ge 1$. Further, for arbitrarily fixed $x,y\in \mathbb{R}$ such that $x\ge y$ and for $t\in [0,1]$ we obtain

$$\begin{array}{rl}|f(t,x)-f(t,y)|& =|\frac{sint}{2}\frac{|x|}{1+|x|}-\frac{sint}{2}\frac{|y|}{1+|y|}|\\ \le \frac{1}{2}|x-y|=\frac{1}{6}\gamma (x-y).\end{array}$$

Thus, the function f satisfies assumption (a₁) with $L=\frac{1}{6}$. It is also easily seen that h is a continuous function. Further, notice that the function k is continuous in $t\in I$ for every $s\in I$ and measurable in $s\in I$ for all $t\in I$ and $k(t,s)\ge 0$. Moreover, we have

$$\begin{array}{rl}{\int }_0^{1}k(t,s)ds& ={\int }_0^1\frac{e^{-s}}{1+t}ds=\frac{1-e^{-1}}{1+t}\\ \le 1-e^{-1}\le \frac{2}{3}=K.\end{array}$$

If we put $\alpha (t)=\frac{3t^2}{4(1+t^4)}$, we have

$$\begin{array}{rl}\alpha (t)& =\frac{3t^2}{4(1+t^4)}\le \frac{t^2}{1+t^4}\\ \le \frac{t^2}{1+t^4}+\frac{1}{27}{\int }_0^1\frac{e^{-s}sint}{2(1+t)}\frac{|\alpha (s)|}{1+|\alpha (s)|}ds\\ =h(t)+\lambda {\int }_0^{1}k(t,s)f(s,\alpha (s))ds.\end{array}$$

This shows that assumption (a₄) holds. Taking $L=\frac{1}{6}$, $K=\frac{2}{3}$ and $\lambda =\frac{1}{27}$, then inequality $L^p{\lambda }^p{K}^p\le \frac{1}{2^{3p-3}}$ appearing in assumption (a₅) has the following form:

$$\frac{1}{6^p}\times \frac{1}{{27}^p}\times \frac{2^p}{3^p}\le \frac{1}{2^{3p-3}}.$$

It is easily seen that each number $p\ge 1$ satisfies the above inequality. Consequently, all the conditions of Theorem 3.1 are satisfied. Hence the integral equation (3.2) has a unique solution in $C(I)$.